Problem: Evaluate the Maclaurin series. $\pi -\frac{{{\pi }^{3}}}{3!}+\frac{{{\pi }^{5}}}{5!}-...+{{\left( -1 \right)}^{n}}\frac{{{\pi }^{2n+1}}}{\left( 2n+1 \right)!}+...$ Choose 1 answer: Choose 1 answer: (Choice A) A $0$ (Choice B) B $1$ (Choice C) C $\cos ( \pi^2 )$ (Choice D) D $-\cos ( \pi^2 )$
Note that the terms of the given series alternate in sign, have only odd factorials in the denominator, and have odd powers of $~\pi~$ in the numerator. This suggests that we work with the Maclaurin series for $~\sin x\,$. Recall the Maclaurin series for $\sin(x)$. $\sin x=x-\frac{{{x}^{3}}}{3!}+\frac{{{x}^{5}}}{5!}-...+{{\left( -1 \right)}^{n}}\frac{{{x}^{2n+1}}}{\left( 2n+1 \right)!}+...$ The given series has $\pi$ instead of $x$ throughout. Hence, the given series converges to $\sin(\pi)$. $\sin(\pi)=0$